Q.Show that the function $f: \mathbb{N} \to \mathbb{N}$, given by $f(1) = f(2) = 1$ and $f(x) = x - 1$, for every $x > 2$, is onto but not one-one.

Concept: Linear Function Properties — checking injectivity and surjectivity for a piecewise-defined function.

Step 1 – One-one (injective)?

For $x > 2$, $f(x) = x-1$ gives distinct outputs for distinct inputs. But $f(1) = 1$ and $f(2) = 1$, so two different inputs map to the same output. Hence $f$ is not one-one.

Step 2 – Onto (surjective)?

We need every $y \in \mathbb{N}$ to have some $x \in \mathbb{N}$ with $f(x) = y$. …

Concept and Intuition

A function $f: A \to B$ is one-one (injective) if different inputs always give different outputs. It is onto (surjective) if every element of the codomain $B$ is actually reached by some input from $A$.

Here, the domain and codomain are both $\mathbb{N} = \{1,2,3,\dots\}$. The function is defined piecewise:

• For $x = 1$ and $x = 2$, the output is $1$.
• For every $x > 2$, the output is $x-1$.

The key observation: the first two values collide at $1$, so injectivity fails immediately. But surjectivity? For any target $n \in \mathbb{N}$, we need to find some $x \in \mathbb{N}$ such that $f(x)=n$. The rule $f(x)=x-1$ for $x>2$ gives us a natural way to hit numbers $2,3,4,\dots$ — just take $x=n+1$. And $n=1$ is already covered by $x=1$ or $x=2$. So every natural number is hit.

Step-by-Step Reasoning

1. Check one-one (injectivity).

A function is one-one if $f(a)=f(b)$ implies $a=b$. Here, $f(1)=1$ and $f(2)=1$. So $f(1)=f(2)$ but $1 \neq 2$. That is a direct counterexample.

Hence $f$ is not one-one.

2. Check onto (surjectivity).

We need to show: for every $n \in \mathbb{N}$, there exists some $x \in \mathbb{N}$ such that $f(x)=n$.

• Case $n=1$: Choose $x=1$ (or $x=2$). Then $f(1)=1$, so $1$ is in the range. …

Method: Deciding one-one / onto for a piecewise-defined function

Use this when $f$ is defined by different rules on different parts of the domain — the special-case pieces are exactly where properties break.

Steps

Step 1: Check one-one across ALL pieces, including collisions between them

Injectivity can fail because two different pieces produce the same output. Compare outputs both within a piece and between pieces; a single collision (e.g. $f(1)=f(2)$) disproves one-one.

Step 2: Check onto by producing a preimage for every target …

Common Mistakes

Mistake 1: Declaring $f$ one-one by looking only at the $x>2$ rule

Why it's wrong: $f(x)=x-1$ is injective on its own, but the special values give $f(1)=f(2)=1$, so two distinct inputs collide and $f$ is not one-one. Correct approach: always test the whole domain, especially the separately-defined values, for injectivity.

Mistake 2: Doubting onto because $f(1)=f(2)=1$ "wastes" an input …